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A whole power of the radicand under Euler's substitution is read as one - #1774
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Rafael-SOWNet merged 1 commit intoOct 5, 2026
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Euler's substitution writes the radicand's powers in its variable as powers of the root, and read a whole power Q^n as the root to the nth, which is Q^(n/2), so it integrated another integrand than the one asked: Q^2/(x + sqrt(Q)) was answered as Q/(x + sqrt(Q)). It reads it as the root to the 2nth now. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Closes #1770.
Euler's substitution read a whole power of the radicand,
Q^n, as the root to thenth power, which isQ^(n/2), and integrated another integrand than the one asked:85d6601c1/((x^2 + 3)^2 (x + sqrt(x^2 + 3)))-2/((sqrt(x^2 + 3) + x)^2 + 3)^21/((x^2 + 2x + 3)^2 (x + sqrt(x^2 + 2x + 3)))x + sqrt(x^2 + 2x + 3)(x^2 + 3)^2/(x + sqrt(x^2 + 3))[0.2, 1.7]it changes by2.08, where the integral is8.48What changes. The substitution writes the radicand's powers in its variable as powers of the root,
Q^(k/2)as the root to thekth; a whole powerQ^nis the root to the2nth now. Read so,Q^2/(x + sqrt(Q))is a rational function past the substitution's bound on the degree, and is declined, as 2.5.0 declined it.Tests:
WholePowerOfTheRadicandUnderEulerTest, the first two rows differentiated back at five points, and the third answered rightly or not at all.The full suite on the head,
efa13a6b, passes, 14,928 tests, and the library builds fornetstandard2.0. The head is one commit on master85d6601c.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura